Question

the equation of the line joining the origin to the point (-4,5)

Answer 1

$\text{Concept used:}$

$\text{The equation of line joining two points (x_1,y_1) and (x_2,y_2) }$

$\text{ is }\;\displaystyle\bf\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\text{Given points are (0,0) and (-4,5)}$

$\text{The equation of line joining two points (0,0) and (-4,5)}$

$\text{ is }\;\displaystyle\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\implies\displaystyle\frac{y-0}{5-0}=\frac{x-0}{-4-0}$

$\implies\displaystyle\frac{y}{5}=\frac{x}{-4}$

$\implies\;\displaystyle\;5x=-4y$

$\implies\;\displaystyle\;5x+4y=0$

$\therefore\textbf{The equation of the line joining origin to the point is 5x+4y=0 }$

Find more:

Point A(7,-3) and B(1,9), find:

a. Slope of AB.

b. Equation of line perpendicular bisector of the line AB.

c. He value of 'p' of (-2,p) lies on it.

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Answer 2

Answer:

 think your question should be:

\text{Prove that:}\displaystyle\;16\;sin10^{\circ}\;sin30^{\circ}\;sin50^{\circ}\;sin70^{\circ}=1Prove that:16sin10∘sin30∘sin50∘sin70∘=1

\text{Consider,}Consider,

\displaystyle\;16\;sin10^{\circ}\;sin30^{\circ}\;sin50^{\circ}\;sin70^{\circ}16sin10∘sin30∘sin50∘sin70∘

=\displaystyle\;16\;sin10^{\circ}(\frac{1}{2})sin50^{\circ}\;sin70^{\circ}=16sin10∘(21)sin50∘sin70∘

=\displaystyle\;8\;sin10^{\circ}\;sin50^{\circ}\;sin70^{\circ}=8sin10∘sin50∘sin70∘

=\displaystyle\;8\;sin(60^{\circ}-10^{\circ})\;sin10^{\circ}\;sin(60^{\circ}-10^{\circ})=8sin(60∘−10∘)sin10∘sin(60∘−10∘)

\text{Using,}Using,

\boxed{\bf\;sin(60^{\circ}-A)\;sinA\;sin(60^{\circ}+A)=\frac{1}{4}sin3A}sin(60∘−A)sinAsin(60∘+A)=41sin3A

=\displaystyle\;8[\frac{1}{4}sin3(10^{\circ})]=8[41sin3(10∘)]

=\displaystyle\;8[\frac{1}{4}sin30^{\circ}]=8[41sin30∘]

=\displaystyle\;8[\frac{1}{4}(\frac{1}{2})]=8[41(21)]

=\displaystyle\;8[\frac{1}{8}]=8[81]

=\displaystyle\;1=1