Question

The line L₁ has equation y = -2x+3.

The line L₂ is perpendicular to L₁ and passes through the point (5, 6).

a.   Find an equation for L₂ in the form ax + by + c = 0, where a, b and c are integers.

The line L₂ crosses the x-axis at the point A and the y-axis at the point B.

b.   Find the x-coordinate of A and the y-coordinate of B.

Given that O is the origin,

c.   Find the area of the triangle OAB.

Solution

a.
 

We are required to find equation of line L₂.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We have coordinates of a point (5,6) on line L₂ but not  slope of the line L₂.

Next we need slope of the line L₂ to write its equation.

We are given that the line L₂ is perpendicular to L₁.

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is; 

Therefore, if we have slope of line L₁, we can find slope of the line L₂.

We do not have slope of both lines L₁ and L₂. However, we are given equation of line L₁ as;

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore;

Hence;

With coordinates of a point on the line L₂ as (5,6) and its slope   at hand, we can write equation of the line L₂.

Point-Slope form of the equation of the line is;

b.
 

We are given that the line L₂ crosses the x-axis at the point A and the y-axis at the point B and we  are required to find the x-coordinate of A and the y-coordinate of B. 

It is evident that we are looking for coordinates of x-intercept (point A) and y-intercept (point B) of  the line L₂. 

First we find coordinates of point A (xintercept).

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line). 

We have found in (a) that equation of the line L₂ is;

We substitute  ;

Hence, x-coordinate of point A is “-7”.

Next we find coordinates of point B (y-intercept)

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line). 

We have found in (a) that equation of the line L₂ is;

We substitute  ;

Hence, y-coordinate of point B is “”.

c.    

We are required to find the area of the triangle OAB.

We draw all these points and subsequent triangle.

Expression for the area of the triangle is;

From the diagram above, we can write;

It is evident from the diagram that OA is the horizontal distance of point A from origin O and that is 7  units. Similarly, OB is the vertical distance of point B from origin O and that is  units. 

Therefore;