MCQ Questions on Coordinate Geometry for Competitive Exams

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.

Area bounded by lines x = 0, y = 0, x + y = 1, 2x + 3y = 6 is
$$\eqalign{ & \Rightarrow 2x + 3y = 6 \cr & \Rightarrow \frac{{2x}}{6} + \frac{{3y}}{6} = 1 \cr & \Rightarrow \frac{x}{3} + \frac{y}{2} = 1 \cr} $$

$$\eqalign{ & {\text{On drawing the lines on graph,}} \cr & \therefore {\text{Area bounded}} \cr & = \frac{1}{2} \times 3 \times 2 - \frac{1}{2} \times 1 \times 1 \cr & = 3 - \frac{1}{2} \cr & = \frac{5}{2} \cr & = 2\frac{1}{2}{\text{ sq}}{\text{. units}} \cr} $$

\[\begin{array}{*{20}{c}} {{\text{Point }}\left( { - 3,\,0} \right){\text{ and }}\left( {0,\,6} \right)} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1}{y_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_2}{y_2}} \end{array}\]
$$\eqalign{ & {\text{Equation of line}} \cr & y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \cr & y - 0 = \frac{{6 - 0}}{{0 - \left( { - 3} \right)}}\left( {x - \left( { - 3} \right)} \right) \cr & y = \frac{6}{3}\left( {x + 3} \right) \cr & y = 2x + 6 \cr} $$

xcosθ - ysinθ = 2 . . . . . . (i)
xsinθ + ycosθ = 4 . . . . . . (ii)
Squaring equation (i) and equation (ii) and adding-
x² + y² = 20

$$\eqalign{ & 3{x_{ \times 3}} - 2{y_{ \times 3}} = {8_{ \times 2}} \to \left( {\text{i}} \right) \cr & \underline {4{x_{ \times 2}} + 3{y_{ \times 2}} = {5_{ \times 2}}} \to \left( {{\text{ii}}} \right) \cr & 17x = 34 \cr & x = 2 \cr & y = - 1 \cr & \left( {2\alpha - \beta } \right) = \left\{ {2\left( 2 \right) - \left( { - 1} \right)} \right\} = 4 + 1 = 5 \cr} $$

$$\eqalign{ & {\text{For no - solution,}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}} \cr & 2x - ky + 7 = 0{\text{ and }}6x - 12y + 15 = 0 \cr & \therefore \frac{2}{6} = \frac{{ - k}}{{12}} \cr & k = 4 \cr} $$

2^x. 2^y = 8
⇒ 2^{x + y} = 2³
On comparing both sides
x + y = 3 . . . . . . . (i)
Also,
9^x. 3^y = 81
⇒ 3^2x. 3^y = 3⁴
⇒ 3^{2x + y} = 3⁴
On comparing both sides
2x + y = 4 . . . . . . . (ii)
On solving equation (i) and (ii)
x = 1
y = 2
∴ (x, y) = (1, 2)

$$\eqalign{ & 2x + 3y = 12 \cr & \Rightarrow \frac{{2x}}{{12}} + \frac{{3y}}{{12}} = 1 \cr & \Rightarrow \frac{x}{6} + \frac{y}{4} = 1 \cr} $$

∴ Straight line 2x + 3y = 12, passes through 1^st, 2^nd and 4^th quadrant.

On plotting the points on graph,

$$\therefore {\text{Area of }}\Delta = \frac{1}{2} \times 5 \times 8 = 20{\text{ sq}}{\text{. units}}$$