A-Level AQA Maths Pure Equation of a Straight Line: The line L has equation 3y

The line L has equation 3y - 4x = 21 The point P has coordinates (15, 2) 5 (a) Find the equation of the line perpendicular to L which passes through P - AQA - A-Level Maths Pure - Question 5 - 2021 - Paper 1

Question 5

The-line-L-has-equation--3y---4x-=-21--The-point-P-has-coordinates-(15,-2)--5-(a)-Find-the-equation-of-the-line-perpendicular-to-L-which-passes-through-P-AQA-A-Level Maths Pure-Question 5-2021-Paper 1.png

The line L has equation 3y - 4x = 21 The point P has coordinates (15, 2) 5 (a) Find the equation of the line perpendicular to L which passes through P. 5 (b) Hen... show full transcript

Worked Solution & Example Answer:The line L has equation 3y - 4x = 21 The point P has coordinates (15, 2) 5 (a) Find the equation of the line perpendicular to L which passes through P - AQA - A-Level Maths Pure - Question 5 - 2021 - Paper 1

Step 1

Find the equation of the line perpendicular to L which passes through P.

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Answer

To find the equation of the line perpendicular to line L, we first need to determine the slope of line L.

Rewrite the equation of line L in slope-intercept form (y = mx + b):

$3y - 4x = 21$

Rearranging gives:

$3y = 4x + 21$

$y = rac{4}{3}x + 7$

From this, we can see that the slope (m) of line L is $rac{4}{3}$ .
Calculate the slope of the perpendicular line: The slope of the line perpendicular to L is the negative reciprocal of the slope of L:

$m_{perpendicular} = - rac{3}{4}$
Using point-slope form: We can use the point-slope form of the line equation, which is given by:

$y - y_1 = m(x - x_1)$

Here, $(x_1, y_1) = (15, 2)$ and $m = - rac{3}{4}$ :

$y - 2 = - rac{3}{4}(x - 15)$
Rearranging to standard form: Rearranging this gives:

$y - 2 = - rac{3}{4}x + rac{45}{4}$

$y = - rac{3}{4}x + rac{53}{4}$

Therefore, the equation of the line perpendicular to L that passes through P is:

$y = - rac{3}{4}x + rac{53}{4}$ .

Step 2

Hence, find the shortest distance from P to L.

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Answer

To find the shortest distance from point P to line L, we can use the formula for the distance D from a point (x_0, y_0) to a line of the form Ax + By + C = 0:

$D = rac{|Ax_0 + By_0 + C|}{ ext{sqrt}(A^2 + B^2)}$

Rearranging the equation of line L:

From the original equation of line L:

$3y - 4x = 21$

We can rewrite it in the standard form:

$4x - 3y + 21 = 0$

Here, A = 4, B = -3, and C = 21.
Substitute P(15, 2) into the distance formula:

Let (x_0, y_0) = (15, 2):

$D = rac{|4(15) - 3(2) + 21|}{ ext{sqrt}(4^2 + (-3)^2)}$

Calculate the numerator:

$= |60 - 6 + 21| = |75| = 75$

Calculate the denominator:

$= ext{sqrt}(16 + 9) = ext{sqrt}(25) = 5$
Final Calculation:

So, the shortest distance D is:

$D = rac{75}{5} = 15$

Hence, the shortest distance from point P to line L is 15.

The line L has equation 3y - 4x = 21 The point P has coordinates (15, 2) 5 (a) Find the equation of the line perpendicular to L which passes through P - AQA - A-Level Maths Pure - Question 5 - 2021 - Paper 1

Worked Solution & Example Answer:The line L has equation 3y - 4x = 21 The point P has coordinates (15, 2) 5 (a) Find the equation of the line perpendicular to L which passes through P - AQA - A-Level Maths Pure - Question 5 - 2021 - Paper 1

Find the equation of the line perpendicular to L which passes through P.

Hence, find the shortest distance from P to L.

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