The length of the segment of the straight lin | vedclass

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

Question

(B) Given points are $(3,3)$ and $(7,6)$.The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.The equation of the line is $(y-3) = \frac{3}{4}(x-

Vedclass · Accepted Answer

$\frac{5}{4}$

Vedclass · Answer

(B) Given points are $(3,3)$ and $(7,6)$.The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.The equation of the line is $(y-3) = \frac{3}{4}(x-3)$,which simplifies to $4y - 12 = 3x - 9$,or $3x - 4y = -3$.To find the $y$-intercept,set $x=0$: $3(0) - 4y = -3 \Rightarrow y = \frac{3}{4}$. The point is $(0, \frac{3}{4})$.To find the $x$-intercept,set $y=0$: $3x - 4(0) = -3 \Rightarrow x = -1$. The point is $(-1, 0)$.The length of the segment between $(0, \frac{3}{4})$ and $(-1, 0)$ is given by the distance formula:$d = \sqrt{(0 - (-1))^2 + (\frac{3}{4} - 0)^2} = \sqrt{1^2 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

The length of the segment of the straight line passing through $(3,3)$ and $(7,6)$ cut off by the coordinate axes is

Similar Questions

$A$ line passes through $(-1, -3)$ and is perpendicular to $x + 6y = 5$. Its $x$-intercept is

Find the slope of the line passing through the points $(3, -2)$ and $(7, -2)$.

The equation of the straight line which passes through the point $(1, -2)$ and cuts off equal intercepts from the axes is:

Find the equation of a straight line that makes an intercept of $3$ units on the negative direction of the $y$-axis and is inclined at an angle of ${\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$ with the $x$-axis.

Find the equation of a straight line which passes through the point $(-1, -1)$ and makes an angle $150^{\circ}$ with the positive direction of the $X$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module