Find the equation of the line joining the points $(- 1, 3)$ and $(4, - 2)$ .

x + y + 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + 4 y - 4 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + y - 2 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $x + y - 2 = 0$
Here, the two points are $(x_{1}, y_{1}) \equiv (- 1, 3)$ and $(x_{2}, y_{2}) \equiv (4, - 2)$
So, the equation of the line in two-point form is :

$y - 3 = \frac{3 - (- 2)}{- 1 - 4} (x + 1)$

$y - 3 = - x - 1$

$x + y - 2 = 0$

Suggest Corrections

Similar questions

The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:

Find the equation of straight line joining the point of intersection of 3x-y+9 = 0 and 2y +x -4 = 0 and the point of intersection of 2x+y = 4 and 2y = x +3.

Q. The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x² − y² −2x + 4y − 3 = 0, is
(a) x² + y² − 2x − 4y + 4 = 0
(b) x² + y² + 2x + 4y − 4 = 0
(c) x² + y² − 2x + 4y + 4 = 0
(d) none of these

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines
$x^{2} - y^{2} - 2 x + 4 y - 3 = 0,$ is

Q. The equation of lines joining (0, 0) and points of intersection

x^{2} + y^{2} + 2 x y = 4, 3 x^{2} + 5 y^{2} - x y = 7

Join BYJU'S Learning Program

Find the equation of the line joining the points (−1,3) and (4,−2).

The correct option is C x+y−2=0Here, the two points are (x1,y1)≡(−1,3) and (x2,y2)≡(4,−2)So, the equation of the line in two-point form is :y−3=3−(−2)−1−4(x+1)y−3=−x−1x+y−2=0

Find the equation of the line joining the points $(- 1, 3)$ and $(4, - 2)$ .

The correct option is C $x + y - 2 = 0$
Here, the two points are $(x_{1}, y_{1}) \equiv (- 1, 3)$ and $(x_{2}, y_{2}) \equiv (4, - 2)$
So, the equation of the line in two-point form is :

$y - 3 = \frac{3 - (- 2)}{- 1 - 4} (x + 1)$

$y - 3 = - x - 1$

$x + y - 2 = 0$