Question

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

1) 7x + 2y = –1
2) 7x + 2y = 1
3) 14x + 4y = –1
4) 14x + 4y = 1

Answer 1

QuesTion :-

→ Line JK passes through the points J(-3,11) and

K(1,-3) .what is the Equation of line JK in standard form is -

Answer :-

→ Option (2) is correct .

$\to \boxed{\rm 2y + 7x = 1}$

To Find :-

→ Equation of line JK .

Approach :-

→ Firstly we will find slope between two given points by the formula -

$\mapsto \boxed{ \rm \: \sf slope = \frac{y_2 - y_1}{x_2 - x_1} }$

And then for finding equation of any line ,we have to take any given point and slope of them and then apply the following formula -

$\to \boxed{\rm \: y - y_1 = m(slope) \{x - x_1 \: \}}$

SoluTion :-

According to the question,

given points are -

$\star \rm (-3 ,11) \to x_1 = - 3 , \:y_1 = 11\\ \star \rm ( 1 , -3) \: \: \to \: x_2 = 1 , \: y_2 = - 3 \\ \\ \rm \mapsto \sf m(slope) = \frac{y_2 - y_1}{x_2 - x_1} \: \\ \\ \mapsto \rm m = \frac{ - 3 - 11}{1 - ( - 3)} \\ \\ \mapsto \rm m = \frac{ - 14}{4} \\ \\ \mapsto \boxed{ \: \rm \: m = - \frac{7}{2} } \\ \\ \rm now \: equation \: of \: line \: passing \: through \\ \rm K(1,-3) \: and \: slope \: (m) = \frac{ - 7}{2} \\ \\ \to \rm y - ( - 3) = \frac{ - 7}{2} (x - 1) \\ \\ \rm \to \: y + 3 = \frac{ - 7}{2} (x - 1) \\ \\ \to \rm 2y + 6 = - 7x + 7 \\ \\ \to \boxed{\rm 2y + 7x = 1}$

Answer 2

(2)$7x+2y=1$

Step-by-step explanation:

Point J(-3,11) and K(1,-3).

$x_1=-3,y_1=11,x_2=1,y_2=-3$

Slope formula:$m=\frac{y_2-y_1}{x_2-x_1}$

Using the formula

Slope of line JK,m=$\frac{-3-11}{1+3}$

Slope of line JK,m=$\frac{-14}{4}=-\frac{7}{2}$

Point-slope form:

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

Using the formula

The equation of line JK passing through the point (1,-3)

$y+3=-\frac{7}{2}(x-1)$

$2y+6=-7x+7$

$7x+2y=7-6=1$

$7x+2y=1$

This is required equation of the line in standard form.

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