P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are the g | vedclass

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

Question

(C) If a line segment $\overline{PQ}$ is parallel to the $Y$-axis,the $x$-coordinates of all points on the line segment must be equal. Therefore,$x_{1

Vedclass · Accepted Answer

$|y_{1}-y_{2}|$

Vedclass · Answer

(C) If a line segment $\overline{PQ}$ is parallel to the $Y$-axis,the $x$-coordinates of all points on the line segment must be equal. Therefore,$x_{1} = x_{2}$. The distance formula between two points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$. Substituting $x_{1} = x_{2}$ into the formula: $PQ = \sqrt{(x_{1}-x_{1})^2 + (y_{2}-y_{1})^2}$ $PQ = \sqrt{0^2 + (y_{2}-y_{1})^2}$ $PQ = \sqrt{(y_{2}-y_{1})^2} = |y_{2}-y_{1}| = |y_{1}-y_{2}|$. Thus,the distance is $|y_{1}-y_{2}|$.

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If $\overline{PQ}$ is parallel to the $Y$-axis,then $PQ = \dots$

Similar Questions

Show that the points $(1,7), (4,2), (-1,-1)$ and $(-4,4)$ are the vertices of a square.

If the distance between $A(-4, -3)$ and $B(6, a)$ is $10$,then $a = \ldots \ldots \ldots . . .$

The distance between the points $A(-4, -6)$ and $B(6, b)$ is $10,$ then $b = \dots$

$A$ person begins his journey from point $A$. From $A$,he moves $800 \, m$ towards north to reach point $B$. From $B$,he moves $500 \, m$ towards east to reach point $C$. From $C$,he moves $400 \, m$ towards north to reach point $D$. Find the direct distance in $m$ from $A$ to $D$.

Do the points $(3,2), (-2,-3)$ and $(2,3)$ form a triangle? If so,name the type of triangle formed.

Vedclass Test Series

Exam Paper Generator

Online Exam Module