The points left( 0, frac{8}{3} right),(1, 3), | vedclass

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

Question

(C) Let the points be $A\left( 0, \frac{8}{3} \right)$,$B(1, 3)$,and $C(82, 30)$.Slope of $AB = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9-8}{3}}{1

Vedclass · Accepted Answer

lie on a straight line.

Vedclass · Answer

(C) Let the points be $A\left( 0, \frac{8}{3} \right)$,$B(1, 3)$,and $C(82, 30)$.Slope of $AB = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9-8}{3}}{1} = \frac{1}{3}$.Slope of $BC = \frac{30 - 3}{82 - 1} = \frac{27}{81} = \frac{1}{3}$.Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A$,$B$,and $C$ are collinear,meaning they lie on a straight line.

The points $\left( 0, \frac{8}{3} \right)$,$(1, 3)$,and $(82, 30)$

Similar Questions

If the lines $ax + 2y + 1 = 0$,$bx + 3y + 1 = 0$,and $cx + 4y + 1 = 0$ are concurrent,then $a, b, c$ are in:

The least positive value of $t$,so that the lines $x=t+\alpha, y+16=0$ and $y=\alpha x$ are concurrent,is

What is the nature of the points $(-a, -b)$,$(0, 0)$,$(a, b)$,and $(a^2, ab)$?

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

The straight lines $4ax + 3by + c = 0$,where $a + b + c = 0$,will be concurrent at the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module