Question

The perpendicular distance of the point p(5,4) from x-axis is

Answer 1

$\rule{400}4$

☆ ANSWER:-

▪︎ Given:-

• A point p(5,4)

▪︎ To find:-

• The perpendicular distance of p from x-axis.

$\rule{400}2$

▪︎ Formula Used:-

• Distance Formula:-

$\tt \: = \sqrt{ ((x \frac{}{2} - x \frac{}{1}) {}^{2} +(y \frac{}{2} - y \frac{}{1}) {}^{2} )}$

$\rule{400}2$

▪︎ Now,

• We know that any point which lies on x-axis must have 0 as its x coordinate.

• Also, here the y coordinate of p is 4 = y coordinate of the point on x-axis because p is the perpendicular to that point.

▪︎ Also, Here:-

$\tt\leadsto \: x \frac{}{1} = 5 \\ \\ \tt\leadsto \: x \frac{}{2} = 0 \\ \\ \tt\leadsto \: y \frac{}{1} = 4 \\ \\ \tt\leadsto \: y \frac{}{2} = 4$

$\rule{400}2$

▪︎ Therefore,

• The perpendicular distance of the point p(5,4) from x-axis,

$\sf = \sqrt{ ((5 - 0) {}^{2} + (4 - 4) {}^{2} )} \\ \\ \sf = \sqrt {((5) {}^{2} + (0) {}^{2} )} \\ \\ \sf = \sqrt {(25 + 0)} \\ \\ \sf = \sqrt{25} \\ \\\sf\huge\pink {\fbox{ = 5.}}$

$\rule{400}2$

${\small{\green{\boxed{\therefore{\bold{The\:Required\:Distance\:=\:5\:Units.}}}}}}$

$\rule{400}4$

(See the above answer in brainly app for better results)

Answer 2

$\rule{400}4</p><p></p><p>☆ ANSWER:-</p><p></p><p>▪︎ Given:-</p><p></p><p>A point p(5,4)</p><p></p><p>▪︎ To find:-</p><p></p><p>The perpendicular distance of p from x-axis.</p><p></p><p>\rule{400}2</p><p></p><p>▪︎ Formula Used:-</p><p></p><p>• Distance Formula:-</p><p></p><p>\tt \: = \sqrt{ ((x \frac{}{2} - x \frac{}{1}) {}^{2} +(y \frac{}{2} - y \frac{}{1}) {}^{2} )}=((x2−x1)2+(y2−y1)2)</p><p></p><p>\rule{400}2</p><p></p><p>▪︎ Now,</p><p></p><p>We know that any point which lies on x-axis must have 0 as its x coordinate.</p><p></p><p>Also, here the y coordinate of p is 4 = y coordinate of the point on x-axis because p is the perpendicular to that point.</p><p></p><p>▪︎ Also, Here:-</p><p></p><p>\begin{gathered}\tt\leadsto \: x \frac{}{1} = 5 \\ \\ \tt\leadsto \: x \frac{}{2} = 0 \\ \\ \tt\leadsto \: y \frac{}{1} = 4 \\ \\ \tt\leadsto \: y \frac{}{2} = 4\end{gathered}⇝x1=5⇝x2=0⇝y1=4⇝y2=4</p><p></p><p>\rule{400}2</p><p></p><p>▪︎ Therefore,</p><p></p><p>• The perpendicular distance of the point p(5,4) from x-axis,</p><p></p><p>\begin{gathered}\sf = \sqrt{ ((5 - 0) {}^{2} + (4 - 4) {}^{2} )} \\ \\ \sf = \sqrt {((5) {}^{2} + (0) {}^{2} )} \\ \\ \sf = \sqrt {(25 + 0)} \\ \\ \sf = \sqrt{25} \\ \\\sf\huge\pink {\fbox{ = 5.}}\end{gathered}=((5−0)2+(4−4)2)=((5)2+(0)2)=(25+0)=25 = 5.</p><p></p><p>\rule{400}2</p><p></p><p>{\small{\green{\boxed{\therefore{\bold{The\:Required\:Distance\:=\:5\:Units.}}}}}}∴TheRequiredDistance=5Units.</p><p></p><p>\rule{400}4</p><p></p><p>(See the above answer in brainly app for better results)</p><p></p><p>$