WMA11 Jan 2019

3.​​ WMA11/01​​ Edexcel IAL P1 January 2019, Q3 (Straight Line Graphs)

The line l1​​ has equation 3x + 5y − 7 = 0​​ 

(a) Find the gradient of l1​​ 

(2)

The line l2​​ is perpendicular to l1​​ and passes​​ through the point (6, −2).​​ 

(b) Find the equation of l2​​ in the form y = mx + c, where m and c are constants.​​ 

(3)

SOLUTION​​ 

a-​​ To find the gradient of the line l1,​​ rearrange the equation to the standard form of linear equation, y=mx+c.​​ 

3x+5y-7=0

5y= -3x+7

y=-35x+75

m1=-35

b-​​ Since the​​ line l1​​ and l2​​ are are perpendicular to each other, the gradient of line l2​​ can be found by using the following equation

m1×m2=-1

-35×m2=-1

m2=53

Now, using the point – slope formula, to find the equation of line l2, where the point through which the line passing is (6, −2) & gradient/slope is 5/3.

y-y1=mx-x1

y=mx+c

y--2=53x-6

y+2=53x-6

3y+6=5x-30

3y=5x-30

3y=5x-36

Hence, the equation of line l2​​ in the form of y=mx+c is​​ 

y=53x -12