Coordinate Geometry - Study Mode

[#66] The graph of the linear equation 3x + 4y = 24 is a straight line intersecting x-axis and y-axis at the points A and B respectively. P(2, 0) and Q$$left( {0,,frac{3}{2}}
ight)$$ xa0are two points on the sides OA and OB respectively of ΔOAB, where O is the origin of the co-ordinate system. Given that AB = 10 cm, then PQ = ?
Correct Answer

(B) 2.5 cm

Explanation

Solution: $$eqalign{
& { ext{By distance formula, }} cr
& { ext{PQ}} = sqrt {{{left( {0 - 2}
ight)}^2} + {{left( {frac{3}{2} - 0}
ight)}^2}} cr
& = sqrt {4 + frac{9}{4}} cr
& = frac{{sqrt {25} }}{2} cr
& = frac{5}{2} cr
& = 2.5{ ext{ cm}} cr} $$

[#67] What is the equation of a line having slope $$ - frac{1}{3}$$ and y-intercept equal to 6?
Correct Answer

(A) x + 3y = 18

Explanation

Solution: $$eqalign{
& left( {y - {y_1}}
ight) = left( {frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}
ight)left( {x - {x_1}}
ight) cr
& Rightarrow left( {y - 6}
ight) = - frac{1}{3} imes left( x
ight) cr
& Rightarrow 3y - 18 = - x cr
& Rightarrow x + 3y = 18 cr} $$

[#68] The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
Correct Answer

(D) -4

Explanation

Solution: $$eqalign{
& ,,,,4x - 2y = 10 cr
& ,,,,4x + ky = 2 cr
& underline {, - ,,,,,, - ,,,,,,,,, - ,,,,} cr
& left( { - k - 2}
ight)y = 8 cr
& { ext{The quardinate of }}y{ ext{ is}} = 4 cr
& left( { - k - 2}
ight) imes 4 = 8 cr
& - k - 2 = 2 cr
& k = - 4 cr} $$

[#69] The slope of the given line is:
Correct Answer

(B) Negative

Explanation

Solution: Slope = tanθ Here, the angle made by the given line is more than 90° ∴ θ > 90° but θ < 180° Since, tanθ in second quadrant is negative. ∴ Slope of the given line is negative.

[#70] The area bounded by the lines x = 0, y = 0, x + y = 1, 2x + 3y = 6 (in square units) is
Correct Answer

(C) $$2frac{1}{2}$$

Explanation

Solution: Area bounded by lines x = 0, y = 0, x + y = 1, 2x + 3y = 6 is $$eqalign{
& Rightarrow 2x + 3y = 6 cr
& Rightarrow frac{{2x}}{6} + frac{{3y}}{6} = 1 cr
& Rightarrow frac{x}{3} + frac{y}{2} = 1 cr} $$ $$eqalign{
& { ext{On drawing the lines on graph,}} cr
& herefore { ext{Area bounded}} cr
& = frac{1}{2} imes 3 imes 2 - frac{1}{2} imes 1 imes 1 cr
& = 3 - frac{1}{2} cr
& = frac{5}{2} cr
& = 2frac{1}{2}{ ext{ sq}}{ ext{. units}} cr} $$