The image of the point $$($$–$$8$$, $$12)$$ with respect to the line mirror $$4x$$ $$+$$ $$7y$$ $$+$$ $$13$$ $$=$$ $$0$$ is

Question

Infinity Learn · Accepted Answer

Equation of the given line $$is4x$$ $$+$$ $$7y$$ $$+$$ $$13$$ $$=$$ $$0$$                                                                    (1)Let$$ Q $$(a$$, $$b) be the image of the point $$P($$–$$8$$, $$12)$$ w.r.t. line (1).Then, PQ ⊥ line (1) and PC $$=$$ CQ.Equation of the line PC $$isy-12=74(x+8)$$[PC is⊥to the line (1) and passes through $$($$–$$8$$, $$12)$$]or $$7x$$ – $$4y$$ $$+$$ $$104$$ $$=$$ $$0$$                                                                (2)Solving$$ Eq. $$(1) and (2), we getx $$=$$ –$$12$$ and y $$=$$ $$5$$.          ∴C ≡ $$($$–$$12$$, $$5)Since$$ C is $$mid-point$$ of PQ,∴$$-12=$$α$$-82$$ and $$5=$$β$$+122$$⇒α$$=-16$$ and β$$=-2$$∴$$Q=(-16$$,$$-2)$$

The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is

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