Question

The line L₁ has equation 4y + 3 = 2x.

The point A (p, 4) lies on L₁.

a.   Find the value of the constant p.

The line L₂ passes through the point C (2, 4) and is perpendicular to L₁.

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

The line L₁ and the line L₂ intersect at the point D.

c.   Find the coordinates of the point D.

d.   Show that the length of CD is .

A point B lies on L₁ and the length of . 

The point E lies on L₂ such that the length of the line CDE = 3 times the length of CD.

e.   Find the area of the quadrilateral ACBE.

Solution

a.
 

We are required to find x-coordinate of the point A (p, 4).

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve  (or the line).

We are given that point A lies on the line L₁ which has equation;

4y + 3 = 2x

We substitute coordinates of point A in the equation of the line L₁;

4(4) + 3 = 2(p)

16 + 3 = 2p

2p =19

b.
 

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 C(2,4) 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, we rearrange the given equation of the line L₁ to write it in slope-intercept form;

Therefore;

Hence;

With coordinates of a point on the line L₂ as C(2,4) and its slope   at hand, we can write
equation of the line L₂.

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

c.
 

We are given that the line L₁ and the line L₂ intersect at the point D and we are required to find the  coordinates of point D.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

Equation of the line L₁ is;

Equation of the line L₂ is;

From equation of the line L₁, we can substitute expression for  in equation of the line L₂;

Single value of y indicates that there is only one intersection point.

With y-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  x-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of y-coordinate of the point of intersection in any of the two equations.

We choose;

Hence coordinates of point D are .

d.    

We are required to find the length of CD.

Expression for the distance between two given points  and is:

Therefore;

We have coordinates of  and . Therefore;

e.
 

We are given that if point B lies on line L₁;

We are also given that the point E lies on L₂ such that the length of the line CDE = 3 times the  length of CD.

We are required to find the area of quadrilateral ACBE.

It is becomes much convenient if we sketch the quadrilateral ACBE as shown below.

First of all we draw the line L₁ passing through point A(9.5,4) and a fictitious point B on the same  line (because we donot have coordinates of point B).

Then we draw the line L₂ which is perpendicular to the line L₁ at point D (taken arbitrarily between  points A and B) where the two lines intersect as given. Here it is quite possible to draw the line L₂ beyond either point A or B outside the segment AB but then it would not make a quadrilateral  ACBE.

Next we sketch points C and E on the line L₂. Then join points A,B,C and E to make quadrilateral  ACBE.

This is shown the figure below.

We are required to find the area of this quadrilateral ACBE.

If we join points A and B, the quadrilateral ACBE is divided into two triangle ABC and ABE. 

Therefore;

Expression for the area of the triangle is;

Since CE (the Line L₂) and AB (the line L₁) are perpendicular, CD and De form the heights of the  two triangles.

We are given that

We have found in (d) that , therefore;