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The point Q has coordinates (2,3)
The point R has coordinates (a, b)
A line perpendicular to QR is given by the equation 5x + 2y = 9
Find an expression for b in terms of a. W

Sagot :

Dammyzeeidn

Answer:

b= (2a+11)/5

Step-by-step explanation:

5x + 2y = 9

2y=9-5x

y= 9/2 - 5x/2

slope, m1= -5/2

for perpendicularity, m1 × m2 = -1

slope for QR= m2=

[tex] = - 1 \div \frac{ - 5}{2} = - 1 \times \frac{2}{ - 5} = \frac{ - 2}{ - 5} = \frac{2}{5} [/tex]

m2 is 2/5

[tex]qr = \frac{b - 3}{a - 2} [/tex]

[tex] \frac{2}{5} = \frac{b - 3}{a - 2} [/tex]

5(b-3)=2(a-2)

5b-15=2a-4

5b=2a-4+15

5b= 2a+11

[tex]b = \frac{2a + 11}{5} [/tex]

Bantssofantasy31idn

Step-by-step explanation:

QR perp = y=-2.5x+4.5

QR gradient = 2/5

y=mx+c

3=(2/5)(2)+c

c=11/5

y=2/5x+11/5

b=2/5a+11/5

USE THIS FOR MATHSWATCH

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