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Q1. The space number space of space points comma space having space both space co minus ordinates space as space integers comma space that space lie space in space the interior space of space the space triangle space with space vertices space left parenthesis 0 comma 0 right parenthesis comma space left parenthesis 0 comma space 41 right parenthesis space and space left parenthesis 41 comma space 0 right parenthesis space is colon
  • 1) 780
  • 2) 861
  • 3) 820
  • 4) 901

Solution

Q2. begin mathsize 12px style If space three space points space left parenthesis straight p comma space 0 right parenthesis comma space left parenthesis straight r comma space straight s right parenthesis space and space left parenthesis 0 comma space straight q right parenthesis space lie space on space line comma space show space that space straight r over straight p plus straight s over straight q equals 1. end style

Solution

Let the given points be D(p, 0), E(r, s) and F(0, q) Since the points lie on the same line,  Slope of DE = Slope of EF                Slope of DE = begin mathsize 12px style fraction numerator straight s space minus space 0 over denominator straight r space minus space straight p end fraction space equals space fraction numerator straight s over denominator straight r space minus space straight p end fraction end style Slope of EF = begin mathsize 12px style fraction numerator straight q space minus space straight s over denominator 0 space minus space straight r end fraction space equals negative space fraction numerator straight q space minus space straight s over denominator straight r end fraction end style
Q3. Without using distance formula, show that the points (-2, 1), (5, 3), (6, 7) and (- 4, 5) are not the vertices of a parallelogram.

Solution

begin mathsize 12px style Let space the space given space points space be space straight A left parenthesis negative 2 comma space 1 right parenthesis comma space straight B left parenthesis 5 comma space 3 right parenthesis comma space straight C left parenthesis 6 comma space 7 right parenthesis space and space straight D left parenthesis negative 4 comma space 5 right parenthesis. Slope space of space AB equals fraction numerator 3 minus 1 over denominator 5 plus 2 end fraction equals 2 over 7 Slope space of space CD equals fraction numerator 5 minus 7 over denominator negative 4 minus 6 end fraction equals fraction numerator negative 2 over denominator negative 10 end fraction equals 1 fifth because space Slope space of space AB not equal to Slope space of space CD comma space AB space is space not space parallel space to space CD. Therefore comma space square ABCD space is space not space straight a space parallelogram. Hence comma space the space given space points space are space not space the space vertices space of space straight a space parallelogram. end style
Q4. Find the slope of line which passes through the point (5, 6) and the mid-point of line joining the points (4, 6) and (-3, 7).

Solution

Mid-point joining the line segment the points (4, 6) and (-3, 7) = Slope of line joining the points (5, 6) and
Q5. If three points (3, 6), (-5, 7) and (x, 1) are collinear, find the value of x.

Solution

If three points are collinear, then
Q6. If the slope of line joining the points (2, 3) and (4, -5) is equal to the slope of line joining the points (x, 5) and (5, 6). Find the value of x.

Solution

Slope of line joining points A (2, 3) and B (4, -5) = Slope of line joining points P (x, 5) and Q (5, 6) = Since Slope of line joining points A (2, 3) and B (4, -5) = Slope of line joining points A (2, 3) and B (4, -5) minus 4 equals fraction numerator 1 over denominator 5 minus x end fraction rightwards double arrow minus 20 plus 4 x equals 1 rightwards double arrow 4 x equals 21 rightwards double arrow x equals 21 over 4
Q7. Find the equations of the lines through the point (3, 2) which are at an angle of 45o with the line x -2y = 3.

Solution

Q8. Locus space of space the space image space of space the space point left parenthesis 2 comma 3 right parenthesis space in space the space line space left parenthesis 2 straight x minus 3 straight y plus 4 right parenthesis plus straight k left parenthesis straight x minus 2 straight y plus 3 right parenthesis equals 0 comma straight k element of straight R comma space is colon
  • 1) Straight line parallel to x-axis
  • 2) circle of radius √2
  • 3) Straight line parallel to y-axis

Solution

Q9. Find the coordinates of the foot of the perpendicular from the point (3, -4) to the line 4x - 15y + 17 = 0.

Solution

The equation of the given line is 4x - 15y + 17 =   0            ...  (i) The equation of a line perpendicular to the given line is 15x + 4y - k = 0, where k is a constant. If this line passes through the point (3, -4), then             15 x 3 + 4 x (-4) - k = 0           45 - 16 - k = 0         k = 29 Therefore the equation of a line passing through the point (3, -4) and perpendicular to the given line is 15x + 4y - 29 = 0                                         ...    (ii)         The required foot of the perpendicular is the point of intersection of lines (i) and (ii). Solving equation (i) and (ii), we get x equals 367 over 241 space a n d space y equals 371 over 241 Therefore, the foot of the perpendicular is given by open parentheses 367 over 241 comma space 371 over 241 close parentheses
Q10. Find the coordinates of point C, which divides the line segment joining the points D (-2, 5) and E (4, 6) in the ratio 2 : 3.

Solution

Let the coordinates of point C is (x, y). Co-ordinates of C =
Q11. If p and q are the lengths of perpendicular from the origin to the lines x cos q - y sin q = k cos 2 q and x sec q + y cosec q = k respectively. Prove that p2 + 4q2 = k2.

Solution

The perpendicular distance from the origin to the line x cos q - y sin q = k cos 2 q                                                   ... (i) Now, x sec q + y cosec q = k fraction numerator x over denominator cos theta end fraction plus fraction numerator y over denominator sin theta end fraction equals k rightwards double arrow x sin theta plus y cos theta equals k sin theta cos theta rightwards double arrow x sin theta plus y cos theta equals k over 2 sin 2 theta   The perpendicular distance q from the origin to the line (ii)
Q12. Show that the line joining the points A(2, -3) and B(-5, 1) is parallel to the line joining the points C(7, -1) and D(0, 3) and perpendicular to the line joining the points P(4, 5) and Q(0, -2).

Solution

begin mathsize 12px style straight A left parenthesis 2 comma space minus 3 right parenthesis space and space straight B left parenthesis negative 5 comma space 1 right parenthesis Slope space of space line space AB equals fraction numerator 1 minus left parenthesis negative 3 right parenthesis over denominator negative 5 minus 2 end fraction equals negative 4 over 7 straight C left parenthesis 7 comma space minus 1 right parenthesis space and space straight D left parenthesis 0 comma space 3 right parenthesis Slope space of space line space CD equals fraction numerator 3 space minus space open parentheses negative 1 close parentheses over denominator 0 minus 7 end fraction equals negative 4 over 7 Since comma space slope space of space line space AB equals slope space of space line space CD comma space AB space parallel to space CD. straight P left parenthesis 4 comma space 5 right parenthesis space and space straight Q left parenthesis 0 comma space minus 2 right parenthesis. Slope space of space line space PQ equals fraction numerator negative 2 minus 5 over denominator 0 minus 4 end fraction equals 7 over 4 Now comma space Slope space of space line space AB cross times slope space of space line space PQ equals negative 4 over 7 cross times 7 over 4 equals negative 1. Hence comma space AB perpendicular PQ. end style
Q13. Find the angle between the line joining the points P (2, 1) and Q (3, -4) and the line joining the points A (-5, 3) and B (6, 7).

Solution

Slope of line joining the points P (2, 1) and Q (3, -4) = m1 = Slope of line joining the points A (-5, 3) and B (6, 7) = m2 = Angle between two lines PQ and AB is: tan theta equals open vertical bar fraction numerator m subscript 2 minus m subscript 1 over denominator 1 plus m subscript 1 m subscript 2 end fraction close vertical bar comma space w h e r e space theta space i s space t h e space a c u t e space a n g l e. rightwards double arrow tan theta equals open vertical bar fraction numerator begin display style 4 over 11 end style minus open parentheses minus 5 close parentheses over denominator 1 plus begin display style 4 over 11 end style cross times minus 5 end fraction close vertical bar equals 59 over 9 rightwards double arrow theta equals tan to the power of minus 1 end exponent open parentheses 59 over 9 close parentheses
Q14. The slope of a line which makes an angle of 60o with the positive direction of y–axis is
  • 1) 1
  • 2) - 1
  • 3) square root of 3
  • 4) fraction numerator 1 over denominator square root of 3 end fraction

Solution

  The angle made by a line with the positive direction of y-axis is complementary angle of the angle made by it with the positive direction of the x-axis. Thus, the angle made with positive direction of x axis is 30o. begin mathsize 12px style therefore space Slope space of space straight a space line equals tan space 30 degree equals fraction numerator 1 over denominator square root of 3 end fraction end style
Q15. The slope of a line segment joining the points G (2a + 3, 4 – 5a) and the point which divides the line segment joining the points E (5, 3) and F (8, 5) in the ratio 3:4 is ½. Find the value of a.

Solution

Let the point D be the point which divides the line segment joining the points (5, 3) and (8, 5) in the ratio 3:4. Therefore the coordinates of point D = Slope of the line joining the points D and G = But the slope of line joining the points D and G is ½.   23 – 14a = 70a – 2 25 = 84a a =
Q16.

Solution

Q17. Find the equation of the lines which cut off intercepts on the axes whose sum and product are 2 and - 8 respectively.

Solution

Let a, b be the intercepts, the lines makes on the axes Sum of intercepts = a + b = 2                                     ---(i) Product of intercepts = ab = -8                                --- (ii) From (i) and (ii), a (2 - a) = -8       2a - a2 = -8         a2 - 2a - 8 = 0         a2 -(4 - 2) a - 8 = 0         (a - 4) (a + 2) = 0          a - 4 = 0 or a + 2 = 0        a = 4 or a = - 2 If a = 4, then 4 + b = 2         b = - 2 And if a = - 2, then             - 2 + b = 2         b = 4. Hence, equations of lines are         x - 2y = 4  and - 2x + y = 4.
Q18. If the lines 3x - 5y + 3 = 0, 5x + 7y - k = 0 and 2x - 3y - 5 = 0 are concurrent, find the value of k.

Solution

Lines are said to be concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the other line. Here given three lines are 3x - 5y + 3 = 0                                    --- (i) 5x + 7y - k = 0                                   ---(ii) And 2x - 3y- 5 = 0                             --- (iii) Solving equation (i) and (iii), we get x = 34 and y = 21. Therefore, the point of intersection of two lines (i) and (iii) is (34, 21). Since above three lines are concurrent, the point (34, 21) will satisfy equation (ii) so that             5 x 34 + 7 x 21- k = 0         170 + 147 - k = 0            k = 317.
Q19. Find the distance between the parallel lines m (x + y) – n = 0 and mx + my + r = 0.

Solution

T h e space p a r a l l e l space l i n e s space a r e space m left parenthesis x plus y right parenthesis minus n equals 0 space a n d space m left parenthesis x plus y right parenthesis plus r equals 0 D i s tan c e space i s space g i v e n space b y space open vertical bar fraction numerator negative n minus r over denominator square root of m squared plus m squared end root end fraction close vertical bar equals fraction numerator n plus r over denominator square root of 2 m squared end root end fraction   open square brackets because space t h e space d i s tan c e space b e t w e e n space t w o space p a r a l l e l space l i n e s space A x plus B y plus C subscript 1 equals 0 space a n d space A x plus B y plus C subscript 2 equals 0 space i s space g i v e n space b y open vertical bar fraction numerator C subscript 1 minus C subscript 2 over denominator square root of A squared plus B squared end root end fraction close vertical bar close square brackets
Q20. Find the new coordinates of the point (4, 7) if origin is shifted to the point (3, 2) by a translation of axes.

Solution

Let new origin (h, k) = (3, 2) and (x, y) = (4, 7) be the given point, therefore new coordinates (X, Y) of (4, 7) are given by  x = X + h and y = Y + k i.e. 4 = X + 3 and 7 = Y + 2 This gives, X = 1 and Y = 5. Thus new coordinates are (1, 5).
Q21. Find the equation of the line perpendicular to the line whose equation is 6x - 7y + 8 = 0 and that passes through the point of intersection of the two lines whose equations are 2x - 3y - 4 = 0 and 3x + 4y - 5 = 0.

Solution

L e t 2 x minus 3 y minus 4 equals 0... left parenthesis i right parenthesis 3 x plus 4 y minus 5 equals 0... left parenthesis i i right parenthesis
Q22. Reduce the equation 4x + 3y – 9 = 0 to the normal form and find their length of perpendicular from origin to the line

Solution

T h e space e q u a t i o n space o f space l i n e space space 4 x plus 3 y minus 9 equals 0 C o m p a r i n g space w i t h space space x cos alpha plus y sin alpha equals p comma space w e space g e t fraction numerator cos alpha over denominator 4 end fraction equals fraction numerator sin alpha over denominator 3 end fraction equals p over 9 rightwards double arrow cos alpha equals fraction numerator 4 p over denominator 9 end fraction space a n d space sin alpha equals fraction numerator 3 p over denominator 9 end fraction space rightwards double arrow cos squared alpha plus sin squared alpha equals fraction numerator 16 p squared over denominator 9 squared end fraction plus fraction numerator 9 p squared over denominator 9 squared end fraction equals fraction numerator 25 p squared over denominator 9 squared end fraction rightwards double arrow 1 equals fraction numerator 25 p squared over denominator 9 squared end fraction rightwards double arrow p equals 9 over 5 N o r m a l space F o r m equals fraction numerator 4 x over denominator 5 end fraction plus fraction numerator 3 y over denominator 5 end fraction equals 9 over 5
Q23. In what ratio, the line joining (-2, 3) and (3, 5) is divided by the line 2x + y = 5.

Solution

The line joining the points (-2, 3) and (3, 5) is divided at P (x1, y1). Let this ratio be k:1.        Point P is This point lies on the line 2x + y = 5.        6k - 4 + 5k + 3 = 5k + 5        6k = 6         k = 1         P divides the line joining the points (-2, 3) and (3, 5) equally
Q24. Find the distance between two parallel lines 2x - y + 3 = 0 And 2x - y - 1 = 0

Solution

Q25.

Solution

T h e space e q u a t i o n space o f space t h e space l i n e space p a s sin g space t h r o u g h space t h e space p o i n t s space left parenthesis 4 comma 4 right parenthesis space a n d space left parenthesis 5 comma 5 right parenthesis space i s space x minus y equals 0 P e r p e n d i c u l a r space d i s tan c e space f r o m space left parenthesis 2 comma 3 right parenthesis space t o space y h e space l i n e space x minus y equals 0 i s open vertical bar fraction numerator 2 minus 3 over denominator square root of 2 end fraction close vertical bar equals fraction numerator 1 over denominator square root of 2 end fraction
Q26. Find the equation of the line passing through the point of intersection of the lines 5x - 2y + 7 = 0 and 3x - 4y - 5 = 0 that has equal intercepts on the axes.

Solution

Let the intercepts on the axes be a and b. Since a = b The point of intersection of lines 5x - 2y + 7 = 0 and 3x - 4y - 5 = 0 is open parentheses minus 19 over 7 comma space minus 23 over 7 close parentheses. But the equation (i) passes through.open parentheses minus 19 over 7 comma space minus 23 over 7 close parentheses minus 19 over 7 minus 23 over 7 equals a rightwards double arrow a equals minus 6 T h e r e f o r e comma space t h e space e q u a t i o n space o f space t h e space l i n e space i s space g i v e n space b y x plus y plus 6 equals 0
Q27. What is the distance between the parallel lines 4(x + y) + 6 = 0 and 8x + 8y + 24 = 0

Solution

T h e space g i v e n space l i n e s space a r e space 4 left parenthesis x plus y right parenthesis plus 6 equals 0 space a n d space 4 left parenthesis x plus y right parenthesis plus 12 equals 0 D i s tan c e equals open vertical bar fraction numerator 12 minus 6 over denominator square root of 4 squared plus 4 squared end root end fraction close vertical bar equals fraction numerator 6 over denominator 4 square root of 2 end fraction equals fraction numerator 3 over denominator 2 square root of 2 end fraction space
Q28. Find the distance of the point (3,4) from the line 3x - 6y - 8 = 0

Solution

Q29. Find the points on the y-axis, whose distances from the line  are 5 units.

Solution

The given line is       8x + 6y = 48 Let any point on the y-axis be (0, y). Perpendicular distance from (0, y) to the given line open vertical bar fraction numerator 6 y minus 48 over denominator square root of 6 squared plus 8 squared end root end fraction close vertical bar equals 5 open vertical bar 6 y minus 48 close vertical bar equals 5 cross times 10 equals 50 T a k i n g space left parenthesis plus right parenthesis space s i g n comma space w e space g e t 6 y minus 48 equals 50 rightwards double arrow y equals 49 over 3 N o w space t a k i n g space left parenthesis minus right parenthesis space s i g n comma space w e space g e t 6 y minus 48 equals minus 50 rightwards double arrow y equals minus 1 third T h e r e f o r e comma space t h e space p o i n t s space a r e space open parentheses 0 comma space 49 over 3 close parentheses space a n d space open parentheses 0 comma space minus 1 third close parentheses space.
Q30. Find the angle between X-axis and the line joining the points (2, -4) and (-3, 5).

Solution

Slope of line joining the points A (2, -4) and B (-3, 5) =     rightwards double arrow space tan theta equals negative 9 over 5 rightwards double arrow theta equals straight pi minus tan to the power of negative 1 end exponent open parentheses 9 over 5 close parentheses


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