Question

Find the equation of the line passing through the intersection of the lines 3x+4y=7 and x-y+2=0 and whose slope is 5

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

35x-7y+18=0

Step-by-step explanation:

Given equations,

3x + 4y = 7 -----(1),

x - y + 2 = 0 ⇒ x = y - 2 -------(2),

From equation (1),

3(y-2) + 4y = 7

3y - 6 + 4y = 7,

7y = 13

⇒ y = 13/7,

From equation (1),

x = 13/7 - 2 = -1/7

So, the intersection point of the given lines is (-1/7, 13/7).

Now, the equation of a line passes through $(x_1, y_1)$ with slope m is,

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

Hence, the equation of the required line having slope 5 is,

$y-\frac{13}{7}=5(x+\frac{1}{7})$

$\frac{7y-13}{7}=\frac{35x+5}{7}$

$7y-13=35x+5$

$\implies 35x-7y+18=0$