Coordinate Geometry - Study Mode
[#16] The line passing through (-3, 4) and (0, 3) is perpendicular to the line passing through (5, 7) and (4, x). What is the value of x?
Correct Answer
(A) 4
Explanation
Solution: Slope (m 1 ) of line which passes through two points (-3, 4) and (0, 3) $$eqalign{
& {m_1} = left( {frac{{3 - 4}}{{0 + 3}}}
ight),,,,,left[ {x08ecause m = frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}
ight] cr
& Rightarrow {m_1} = frac{{ - 1}}{3} cr} $$ Similarly, slope (m 2 ) of line which passes through the two points (5, 7) and (4, x) $${m_2} = frac{{x - 7}}{{4 - 5}} = - left( {x - 7}
ight)$$ ∵ These lines perpendicular to each other, $$eqalign{
& herefore {m_1} imes {m_2} = - 1 cr
& frac{{ - 1}}{3} imes left[ { - left( {x - 7}
ight)}
ight] = - 1 cr
& x - 7 = - 3 cr
& x = 4 cr} $$
[#17] If (2, 0) is a solution of the linear equation 2x + 3y = 5, then the value of k is:
Correct Answer
(D) 4
Explanation
Solution: Given that (2, 0) is solution of equation 2x + 3y = k ∴ On putting value of x & y in the above equation, 2 × 2 + 3 × 0 = k k = 4
[#18] What is the y-intercept of the linear equation 59x + 14y - 112 = 0?
Correct Answer
(A) 8
Explanation
Solution: $$eqalign{
& 59x + 14y - 112 = 0 cr
& Rightarrow 59x + 14y = 112 cr
& Rightarrow frac{{59x}}{{112}} + frac{{14y}}{{112}} = 1 cr
& Rightarrow frac{x}{{frac{{112}}{{59}}}} + frac{y}{8} = 1 cr
& herefore y{ ext{ - intercept of the line is }}8 cr} $$
[#19] The graphs of the equations 2x + 3y = 11 and x - 2y + 12 = 0 intersects at P(x 1 , y 1 ) and the graph of the equation x - 2y + 12 = 0 intersects the x-axis at Q(x 2 , y 2 ). What is the value of (x 1 - x 2 + y 1 + y 2 )?
Correct Answer
(C) 15
Explanation
Solution: $$eqalign{
& 2x + 3y = 11{ ext{ }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}left( { ext{i}}
ight) cr
& x - 2y = - 12{ ext{ }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{left( {{ ext{ii}}}
ight)_{ imes 2}} cr
& ,2x - 4y = - 24 cr
& underline {,2x + 3y = 11,} cr
& ,,,,,,,,,,,,7y = 35 cr
& ,,,,,,,,,,,,,,y = 5 cr
& { ext{From equation}}left( { ext{i}}
ight) cr
& 2x + 15 = 11 cr
& x = - 2 cr
& left( {{x_1},,{y_1}}
ight) = left( { - 2,,5}
ight) cr
& { ext{At }}x{ ext{ - axis}},,y = 0 cr
& x - 2 imes 0 = - 12 cr
& x = - 12 cr
& left( {{x_2},,{y_2}}
ight) = left( { - 12,,0}
ight) cr
& {x_1} - {x_2} + {y_1} + {y_2} = - 2 + 12 + 5 + 0 = 15 cr} $$
[#20] A line passing through the origin perpendicularly cuts the line 3x - 2y = 6 at point M. Find the co-ordinates of M.
Correct Answer
(B) $$left( {frac{{18}}{{13}},,frac{{ - 12}}{{13}}}
ight)$$
Explanation
Solution: Equation of line perpendicular to line 3x - 2y = 6 is 2x + 3y + p = 0 As, the line passes through origin (0, 0) ∴ 2 × 0 + 3 × 0 + p = 0 ∴ p = 0 Now, the equation is 2x + 3y = 0 ∴ 2x = -3y x = $$frac{{ - 3}}{2}$$ y By putting this value in equation 3x - 2y = 6 $$eqalign{
& Rightarrow 3left( {frac{{ - 3}}{2}}
ight)y - 2y = 6 cr
& Rightarrow frac{{ - 9}}{2}y - 2y = 6 cr
& Rightarrow frac{{ - 13y}}{2} = 6 cr
& Rightarrow y = - frac{{12}}{{13}} cr
& herefore x = frac{{ - 3}}{2} imes left( {frac{{ - 12}}{{13}}}
ight) cr
& x = frac{{18}}{{13}} cr
& herefore { ext{Co - ordinate of point}} cr
& { ext{M}} = left[ {frac{{18}}{{13}},, - frac{{12}}{{13}}}
ight] cr} $$