(A) The slope $m$ of the line passing through $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$ is given by:$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The slope $m$ of the line passing through $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$ is given by:$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}$.Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(-1, 1)$:$y - 1 = -\frac{5}{3}(x - (-1))$$3(y - 1) = -5(x + 1)$$3y - 3 = -5x - 5$$5x + 3y + 2 = 0$.

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

Easy

Find the equation of the line passing through the points $(-1, 1)$ and $(2, -4)$.

A
$5x + 3y + 2 = 0$
B
$5x + 3y - 2 = 0$
C
$5x + 3y + 1 = 0$
D
$5x + 3y - 1 = 0$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Pair of straight lines

Subtopic

Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

The lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 3$ and the line $x + y = 2$ are:

Difficult

View Solution

The lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ will be equidistant from the origin,if

View Solution

Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$. If $\angle AOB=90^{\circ}$,then a value of $\lambda$ is

MediumAP EAMCET 2022

View Solution

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

DifficultAP EAMCET 2014

View Solution

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

MediumAP EAMCET 2019

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Find the equation of the line passing through the points $(-1, 1)$ and $(2, -4)$.

Similar Questions

The lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 3$ and the line $x + y = 2$ are:

The lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ will be equidistant from the origin,if

Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$. If $\angle AOB=90^{\circ}$,then a value of $\lambda$ is

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

Vedclass Test Series

Exam Paper Generator

Online Exam Module