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Parin Sir > Quantitative Aptitude (Maths) > Co–Ordinate Geometry > Examples 11 to 15 of Co–Ordinate Geometry > Examples 11 to 15 of Co–Ordinate Geometry

Examples 11 to 15 of Co–Ordinate Geometry

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  1. In a rectangle ABCD, if B (3, 1) and D (7, 5) are the endpoints of diagonal \overline{\mathrm{BD}}. If slope of  \overline{\mathrm{AC}} = 2, then find the equation of  \overline{\mathrm{AC}}.

 Soln. In a rectangle, mid–point of  \overline{\mathrm{AC}} =   mid–point of  \overline{\mathrm{BD}}

=\left(\frac{3+7}{2},\ \frac{1+5}{2}\right)  =   (5, 3)

 Hence, \overline{\mathrm{AC}}  passes through (5, 3) with the slope of  2.

Equation of  \overline{\mathrm{AC}}  is  :  y  –  y1  =  m (x – x1)

∴  y – 3  =  2 ( x – 5 )   ⇒  2x – y – 7 = 0.

  1. Find the equation of a line passes through (1, – 2) and cuts equal intercepts on both the axes.

 Soln. Equation of  a line which makes intercepts on respective axes is : \frac{x}{a} + \frac{y}{b} = 1

Here, the required line cuts equal intercepts on both the axes  ⇒  a = b

∴ Equation of  a required line is  :  x  +  y  =  a.

Which passes through (1, – 2)  ⇒  1  – 2  =  a  ⇒  a = – 1

Hence, equation of  a required line is  :  x  +  y  =  – 1.

  1. Find both the intercepts and slope for the line 3x + 5y – 7 = 0.

Soln. If we compare 3x + 5y – 7 = 0 with ax + by + c = 0, we get a = 3, b = 5 and c = – 7.

Now, X – intercept =  –\frac{c}{a} = \frac{7}{3},           Y – intercept  =  –\frac{c}{b} = \frac{7}{5}

And Slope  =  m  =  -\frac{a}{b} = -\frac{3}{5}

  1. If the slope of the curve in a graph is constant, then the curve is ………..

Soln. The curve is line.

  1. Find the length of perpendicular from (1, 3) to 5y – 12 y – 11 =  0.

 Soln. Length of  a perpendicular  =  p  =  \frac{\left|{\rm ax}_1+\mathrm{b} \mathrm{y}_1+c\right|}{\sqrt{a^2+b^2}} =  \frac{\left|5(1)-12(3)-11\right|}{\sqrt{5^2+12^2}} = \frac{42}{13}

 

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