Question

Write an equation that passes through the point (1, 3) and is perpendicular to x – 2y = -8.

Answer 1

y=½x+5

Explanation:

Passes through, (1,3)

Perpendicular to x-2y=-8.

Step 1

write the equation in form of y=mx+c

x-2y=8

-2y=-x+8

Dividing both sides of the equation by -2 we get;

y=-x/2+8

y=-½x+8

Thus the gradient is -½. Call it M1

Recall, The product of the gradients(slopes) of perpendicular lines is -1 i.e

M1 × M2= -1

Solve for M2

-½×M2=-1.

Dividing both sides by -½ we get

M2=-1×-2

M2=2

Now, use this slope(M2) with the points (1,3) to find the new equation that is perpendicular to x-2y=-8

Note: Given two points (x1,y1) and (x2,y2) the the slope would be ∆x/∆y i.e (x1-x2)/(y1-y2).

Therefore, identify new point (x,y). Use this new point with point (1,3) and the slope M2 to get the new equation.

Slope=∆x/∆y

M2=(x-1)/(y-3)

but M2 is 2

2=(x-1)/(y-3)

2/1=(x-1)/(y-3) Note: 2= 2/1

Cross multiplying we get:

2(y-3)=1(x-1)

Open the brackets

2y-6=x-1

Solve for y so as to write the equation in form of y=mx+c

2y=x-1 +6

2y=x+5

Divide both sides of the equation by 2

y=x/2+5

y=½x+5

Answer 2

y = -2x + 5

Explanation:

Given:

Passes through point (1, 3)

Perpendicular to x – 2y = -8

Solve:

x – 2y = -8

y = 1/2x + 4

The slope is m = 1/2

The slope of the perpendicular line is the inverse of the slope of the original equation.

The slope of the inverse equation is m = -2.

Making an inverse equation of y = -2x + a

Find a:

Use point, (1, 3) where (x, y):

3 = (-2)*(1) + a

a = 5

y = -2x + 5