The straight line L has the equation
$3y = 4x + 7$
The point A has coordinates $(3, -5)$ - Edexcel - GCSE Maths - Question 17 - 2019 - Paper 2
Question 17
The straight line L has the equation
$3y = 4x + 7$
The point A has coordinates $(3, -5)$.
Find an equation of the straight line that is perpendicular to L and p... show full transcript
Worked Solution & Example Answer:The straight line L has the equation
$3y = 4x + 7$
The point A has coordinates $(3, -5)$ - Edexcel - GCSE Maths - Question 17 - 2019 - Paper 2
Step 1
Identify the gradient of L
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Answer
First, we need to rewrite the equation of the line L in slope-intercept form (y = mx + b). Starting with: 3y=4x+7
Dividing through by 3 gives: y = rac{4}{3}x + rac{7}{3}
From this form, we can see that the gradient (slope) of line L is rac{4}{3}.
Step 2
Determine the gradient of the perpendicular line
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Answer
The gradient of a line that is perpendicular to another is the negative reciprocal of the original line's gradient. Therefore, the gradient (m) of the perpendicular line is: m = - rac{1}{ rac{4}{3}} = - rac{3}{4}.
Step 3
Find the equation of the line through point A
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Answer
Now we can use the point-slope form of the equation of a line, which is: y−y1=m(x−x1)
Using point A (3, -5) as (x1,y1) and the calculated gradient, we substitute: y - (-5) = - rac{3}{4}(x - 3)
This simplifies to: y + 5 = - rac{3}{4}x + rac{9}{4}.
Finally, rearranging gives: y = - rac{3}{4}x + rac{9}{4} - 5 = - rac{3}{4}x - rac{11}{4}.
Thus, the equation of the straight line that is perpendicular to L and passes through point A is: y = - rac{3}{4}x - rac{11}{4}.
