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BECE Mathematics 2020 Paper

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Question : 30

Total: 40

Find the equation of the straight line passing through the points

(-3,5)

(6,8)

$y=\frac{1}{3^x}$
$y=\frac{1}{3^x}+6$
$y=\frac{1}{3^x}-10$
$y=\frac{1}{3^x}+14$

Solution:

Concept:

For any straight line, the slope-intercept form is

y=mx+c

, where

m

is the gradient and

c

is the

y

-intercept.

Explanation:

Let

(x_1,y_1)=(-3,5)

and

(x_2,y_2)=(6,8)

.

The gradient is

m=\frac{y_2-y_1}{x_2-x_1}=\frac{8-5}{6-(-3)}=\frac{3}{9}=\frac{1}{3}

.

Substitute

m=\frac{1}{3}

and the point

(-3,5)

into

y=mx+c

.

5=\frac{1}{3}(-3)+c

\Rightarrow 5=-1+c

\Rightarrow c=6

Verify with the other point

(6,8)

.

8=\frac{1}{3}(6)+c=2+c

\Rightarrow c=6

Answer:

The equation of the required line is

y=\frac{1}{3}x+6

.

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