Coordinate Geometry - Study Mode

[#36] Find k, if the line 4x - y = 1 is perpendicular to the line 5x - ky = 2?
Correct Answer

(B) -20

Explanation

Solution: $$eqalign{
& { ext{Given,}} cr
& 4x - y = 1 cr
& herefore y = 4x - 1 cr
& 5x - ky = 2 cr
& herefore y = frac{{5x}}{k} - frac{2}{k} cr
& y = {m_1}x + C cr
& {m_1} = frac{5}{k} cr} $$ If the lines are ⊥ then the product of their slope is -1 [x08egin{array}{l}
{m_1} imes {m_2} = - 1\
4 imes frac{5}{k} = - 1,,,,,,,,,,left{ x08egin{array}{l}
{m_1} = 4\
{m_2} = frac{5}{k}
end{array}
ight.\
herefore k = - 20
end{array}]

[#37] At what point does the line 2x - 3y = 6 cuts the Y axis?
Correct Answer

(D) (0, -2)

Explanation

Solution: 2x - 3y = 6 Cut Y-axis at x = 0 ∴ 0 - 3y = 6 $$x08oxed{y = - 2}$$ Hence, point will be (0, -2)

[#38] The line passing through (-2, 5) and (6, b) is perpendicular to the line 20x + 5y = 3. Find b?
Correct Answer

(C) 7

Explanation

Solution: Equation of given line $$eqalign{
& Rightarrow 20x + 5y = 3 cr
& Rightarrow 5y = 3 - 20x cr
& Rightarrow y = - 4x + frac{3}{5} cr} $$ Slope of line, m 1 = -4 If two are lines are ⊥ then the product of their slope = -1 $$eqalign{
& {m_1} imes {m_2} = - 1 cr
& - 4 imes {m_2} = - 1 cr
& {m_2} = frac{1}{4} cr} $$ Lines passing through the points (-2, 5) and (6, b) Therefore, $$eqalign{
& { ext{Slope}} = frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = frac{{b - 5}}{{6 + 2}} = frac{{b - 5}}{8} cr
& { ext{According to the question,}} cr
& frac{{b - 5}}{8} = frac{1}{4} cr
& b - 5 = 2 cr
& herefore b = 7 cr} $$

[#39] Find k, if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = -4?
Correct Answer

(D) 2

Explanation

Solution: $$eqalign{
& 2x - 3y = 11 cr
& 3y = 2x - 11 cr
& y = frac{2}{3}x - frac{{11}}{3} cr
& left( {y = mx + c,{ ext{ where }}m{ ext{ is slope}}}
ight) cr
& { ext{Slope}} = {m_1} = frac{2}{3} cr
& 3x + ky = - 4 cr
& ky = - 3x - 4 cr
& y = - frac{3}{k}x - frac{4}{k} cr
& {m_2} = - frac{3}{k} cr} $$ Relation between slope of perpendicular lines $$eqalign{
& {m_1}{m_2} = - 1 cr
& Rightarrow left( {frac{2}{3}}
ight) imes left( { - frac{3}{k}}
ight) = 1 cr
& Rightarrow k = 2 cr} $$

[#40] A point in the 4 th quadrant is a unit away from x-axis and 7 unit away from y-axis. The point is at:
Correct Answer

(A) (7, -6)

Explanation

Solution: The point in 4 th quadrant that is 6 unit away from x-axis and 7 unit away from y-axis is (7, -6).