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Question Number 68237 by smartsmith459@gmail.com last updated on 07/Sep/19

1. find the equation of the line making an angle of 135^(° ) with O_(x ) and passing through thepoints?(−2,5) 2. find the slope of the line through the points (5,3)and (7,2). find (i) the perpendicular form (ii) find the intercept form of its equation. 3. Determine the gradient of the straight line graph passing through the co-ordinates: (i) (2,7) and (−3,4) (ii) (((1 )/4), ((-3)/4)) and (((-1)/2), (5/8)).

$$\mathrm{1}.\:{find}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{making}\:{an}\:{angle}\:{of}\:\mathrm{135}^{°\:} {with}\:{O}_{{x}\:} \:{and}\:{passing}\:{through}\:{thepoints}?\left(−\mathrm{2},\mathrm{5}\right) \\ $$$$\mathrm{2}.\:{find}\:{the}\:{slope}\:{of}\:{the}\:{line}\:{through}\:{the}\:{points}\:\left(\mathrm{5},\mathrm{3}\right){and}\:\left(\mathrm{7},\mathrm{2}\right).\:{find}\:\left({i}\right)\:{the}\:{perpendicular}\:{form}\:\left({ii}\right)\:{find}\:{the}\:{intercept}\:{form}\:{of}\:{its}\:{equation}. \\ $$$$\mathrm{3}.\:{Determine}\:{the}\:{gradient}\:{of}\:{the}\:{straight}\:{line}\:{graph}\:{passing}\:{through}\:{the}\:{co}-{ordinates}: \\ $$$$\left({i}\right)\:\left(\mathrm{2},\mathrm{7}\right)\:{and}\:\left(−\mathrm{3},\mathrm{4}\right) \\ $$$$\left({ii}\right)\:\left(\frac{\mathrm{1}\:}{\mathrm{4}},\:\frac{-\mathrm{3}}{\mathrm{4}}\right)\:{and}\:\left(\frac{-\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{5}}{\mathrm{8}}\right). \\ $$

Commented by kaivan.ahmadi last updated on 07/Sep/19

1. m=tan135=tan(90+45)=−cot45=−((√2)/2) y−5=−((√2)/2)(x+2)⇒y=−((√2)/2)x+(5−(√2)) 2. m=((3−2)/(5−7))=−(1/2) y−3=−(1/2)(x−5)⇒y=−(1/2)x+((11)/2) perpendicular slope m_1 =2 3. (i). y−7=((4−7)/(−3−2))(x−2)⇒y=(3/5)x+((29)/5)⇒ f(x,y)=3x−5y+29=0 ▽f=(3,−5)

$$\mathrm{1}. \\ $$$${m}={tan}\mathrm{135}={tan}\left(\mathrm{90}+\mathrm{45}\right)=−{cot}\mathrm{45}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${y}−\mathrm{5}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}+\mathrm{2}\right)\Rightarrow{y}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{x}+\left(\mathrm{5}−\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{2}. \\ $$$${m}=\frac{\mathrm{3}−\mathrm{2}}{\mathrm{5}−\mathrm{7}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}−\mathrm{3}=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{5}\right)\Rightarrow{y}=−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{11}}{\mathrm{2}} \\ $$$${perpendicular}\:{slope}\:\:{m}_{\mathrm{1}} =\mathrm{2} \\ $$$$\mathrm{3}. \\ $$$$\left({i}\right). \\ $$$${y}−\mathrm{7}=\frac{\mathrm{4}−\mathrm{7}}{−\mathrm{3}−\mathrm{2}}\left({x}−\mathrm{2}\right)\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{5}}{x}+\frac{\mathrm{29}}{\mathrm{5}}\Rightarrow \\ $$$${f}\left({x},{y}\right)=\mathrm{3}{x}−\mathrm{5}{y}+\mathrm{29}=\mathrm{0} \\ $$$$\bigtriangledown{f}=\left(\mathrm{3},−\mathrm{5}\right) \\ $$$$ \\ $$

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