Question

Show that the lines x-7y-22=0,3x+4y+9=0and 7x+y-54=0 form a right angled triangle

Answer 1

Answer:

$\huge\fbox{☘Answer}$

i) θ=-π4

∴ Slope of the line=m=tanθ⇒Slope of the line=tan-π4=-tanπ4=-1

Hence, the slope of the line is -1.

(ii) θ=2π3

∴ Slope of the line=m=tanθ⇒Slope of the line=tan2π3=-tanπ3=-3

Hence, the slope of the line is -3.

(iii) θ=3π4

∴ Slope of the line=m=tanθ⇒Slope of the line=tan3π4=-tanπ4=-1

Hence, the slope of the line is -1.

(iv) θ=π3

∴ Slope of the line=m=tanθ⇒Slope of the line=tanπ3=3

Answer 2

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: prove : \\ △ABC \: is \: a \: right \: angled \: triangle \\ \\ so \: let \: here \: accordingly \\ x - 7y = 22 \: , \: 3x + 4y = - 9 \: \: and \: \\ 7x + y = 54 \: \: be \: resp ective \: equations\\of \: AB \: ,BC \: \: and \: AC \\ \\ so \: thus \: then \\ comparing \: the \: equation \\ x - 7y = 22 \: \: with \: \: ax + by = c \\ we \: get \\ a = 1 \\ b = - 7 \\ c = - 22 \\ \\ we \: have \: \\ m1 = - \frac{a}{b} \\ \\ = - \frac{1}{( - 7)} \\ \\ slope \: of \: AB = m1= \frac{1}{7} \: \: \: \: (1) \\ \\ similarly \: \: comparing \: the \: equation \\ 7x + y = 54 \: \: with \: \: ax + by = c \\ we \: get \\ a = 7 \\ b = 1 \\ c = - 54 \\ \\ thus \: then \\ m2 = - \frac{a}{b} \\ \\ = - \frac{7}{1} \\ \\ slope \: of \: AC = m2= - 7 \: \: \: \: (2)$

$now \: applying \\ (1) \times (2) \\ we \: get \\ \\ m1.m2 = \frac{1}{7} \times ( - 7) \\ \\ m1.m2= - 1 \\ \\ since \: here \\ product \: of \: slopes \: of \: \\ AB \: and \: AC \: is \: \: - 1 \\ AB \: and \: AC \: are \: perpendicular \\ to \: each \: other \\ \\ thus \: then \\ by \: converse \: of \: pythagoras \: theorem \\ we \: conclude \: that \\ △ABC \: is \: a \: right \: angled \: triangle \\ right \: angled \: at \: A$