Question

L1 and L2 are two straight lines. The origin of the coordinate axes is O. L1 has equation 5x + 10y = 8 L2 is perpendicular to L1 and passes through the point with coordinates (8, 6) L2 crosses the x-axis at the point A. L2 intersects the straight line with equation x = –3 at the point B. Find the area of triangle AOB. Show your working clearly.

Answer 1

Answer:

40 sq units

Step-by-step explanation:

$5x + 10y = 8\\\Rightarrow y=\dfrac{8-5x}{10}\\\Rightarrow y=-\dfrac{1}{2}x+\dfrac{2}{5}$

Slope of $L_2$ is $m_2$

$m_1m_2=-1\\\Rightarrow m_2=-\dfrac{1}{m_1}\\\Rightarrow m_2=-\dfrac{1}{-\dfrac{1}{2}}\\\Rightarrow m_2=2$

$L_2$ passes through point $(8,6)$

$y-6=2(x-8)\\\Rightarrow y=2x-16+6\\\Rightarrow y=2x-10$

Point at x axis where $L_2$ intersects is

$0=2x-10\\\Rightarrow x=\dfrac{10}{2}\\\Rightarrow x=5$

Point $A$ of the triangle will be $(5,0)$

$L_2$ intersects line $x=-3$. The point is

$y=2\times(-3)-10\\\Rightarrow y=-6-10\\\Rightarrow y=-16$

The points of the triangle are $A(5,0), O(0,0), B(-3,-16)$

Area of triangle is given by

$A=\dfrac{|A_x(B_y-O_y)+B_x(A_y-O_y)+O_x(A_y-B_y)|}{2}\\\Rightarrow A=\dfrac{|5(-16-0)+-3(0-0)+0(0-(-16))|}{2}\\\Rightarrow A=40\ \text{sq units}$

The area of the triangle is 40 sq units.