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# A straight line passes through a fixed point (h, k). Then the locus of the feet of the perpendiculars on it from the origin is

A + -hx-ky=0
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B -hx+ky=0
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C +hx+ky=0
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D + -hx-ky=0
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Solution

## The correct option is A + -hx-ky=0Let the equation of the straight line be xcosα+ysinα–p=0…..(i) Since, the straight line (i) passes through the point (h, k), therefore hcosα+ksinα–p=0…..(ii) On substracting Eq. (ii) from Eq. (i), we get (x−h)cosα+(y–k)sinα=0…..(iii) Now, the equation of the straight line which is perpendicular to line (i) and passes through the origin, is xsinα–ycosα=0…..(iv) The required locus will be obtained by eliminating α from the Eqs. (iii) and (iv). From Eq. (iv), xsinα=ycosα ⇒xcosα=ysinα⇒xcosα=ysinα=√x2+y2√cos2α+sin2α=√x2+y2⇒x(x−h)+y(y−k)=0⇒x2−hx+y2−ky=0∴x2+y2−hx−ky=0 Which is the required locus. Hence, (a) is the correct answer.  Suggest Corrections  0      Similar questions  Related Videos   Straight Line
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