Question

A line passes through the points (1, 4) and (5,8). A second line passes through the points (2, 10) and (6, 4). At what point do the two lines intersect? ​

Answer 1

Answer: x = 3.67, y = 6.67

Step-by-step explanation:

Find gradient of both lines:

gradient formula = y2 - y1 divided by x2 - x1

first line = (8-4)/ (5-1) = 1

second line gradient = (6-10)/(4-2) = -2

First line equation: y = x +c

Find c by inputting a point on the line ->

4 = 1 +c

c = 3

equation ; y = x + 3

Second line: y = -2x (gradient)+ c

10 = -4 + c

c = 14

equation; y = -2x + 14

x+3 = -2x + 14

3x = 11

x = 11/3 -> 3.67

Y = 3.67 + 3 = 6.67

Answer 2

Answer:

(4, 7 )

Step-by-step explanation:

The first step is to obtain the equations of the lines and then solve simultaneously.

The equation of a line in slope0 intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 4) and (x₂, y₂ ) = (5, 8)

m = $\frac{8-4}{5-1}$ = $\frac{4}{4}$ = 1 then

y = x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (1, 4 ) , then

4 = 1 + c ⇒ c = 4 - 1 = 3

y = x + 3 → (1)

Repeat for points on line 2 (2, 10) and (6, 4)

m = $\frac{4-10}{6-2}$ = $\frac{-6}{4}$ = - $\frac{3}{2}$

y = - $\frac{3}{2}$ x + c

Using (2, 10 ) to find c

10 = - 3 + c ⇒ c = 10 + 3 = 13

y = - $\frac{3}{2}$ x + 13 → (2)

Equate the right sides of (1) and (2)

x + 3 = - $\frac{3}{2}$ x + 13 ( multiply through by 2 to clear the fraction )

2x + 6 = - 3x + 26 ( add 3x to both sides )

5x + 6 = 26 ( subtract 6 from both sides )

5x = 20 ( divide both sides by 5 )

x = 4

Substitute x = 4 into (1) for corresponding value of y

y = 4 + 3 = 7

point of intersection = (4, 7 )