Straight Lines Set 2

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Example –3

Prove that the centroid G of a triangle divides the line joining its circumcentre C and its orthocentre H in the ratio 1 : 2.

Solution: To make our task simpler, we choose a coordinate frame in which the triangle’s vertices have coordinates that are “easy” to work with. One such choice is shown below.

Now, we find G, C and H :

The centroid \(G({x_G},\,\,{y_G})\) :

\[\begin{align}&{x_G} = \frac{{ - a + a + b}}{3} = \frac{b}{3} \\ &{y_G} = \frac{{0 + 0 + c}}{3} = \frac{c}{3}  \\   \Rightarrow \qquad &G \equiv \left( {\frac{b}{3},\,\,\frac{c}{3}} \right)  \\ \end{align} \]

The circumcentre: To find C, we need two perpendicular bisectors \(( \bot B).\)

The \(\bot B\) of PQ is

\(x = 0\)

Since the slope of PR is \(\begin{align}\frac{c}{{b + a}},\end{align}\)  the equation of the \( \bot B\) of PR is

\[y - \frac{c}{2} = \left( { - \frac{{b + a}}{c}} \right)\left( {x - \frac{{b - a}}{2}} \right)\]

The two \(\bot Bs\) intersect at C:

\[C \equiv \left( {0,\,\,\frac{{{b^2} - {a^2} + {c^2}}}{{2c}}} \right)\]

The orthocentre \(H({x_H},\,{y_H})\) : The altitude from R onto PQ is simply 

\(x = b\)

Let find the altitude from Q onto PR:

\[y - 0 = \left( { - \frac{{b + a}}{c}} \right)(x - a)\]

The two altitudes intersect at

\[H \equiv \left( {b,\,\, - \frac{{{b^2} - {a^2}}}{c}} \right)\]

The point which divides CH in the ratio 1 : 2 is

\[\begin{align}\left( {\frac{{1 \times b + 2 \times 0}}{3},\,\,\frac{{1 \times \left( { - \frac{{{b^2} - {a^2}}}{c}} \right) + 2 \times \left( {\frac{{{b^2} - {a^2} + {c^2}}}{{2c}}} \right)}}{3}} \right) \equiv \left( {\frac{b}{3},\,\,\frac{c}{3}} \right)\end{align}\]

which is the same as G. Thus, G divides CH in the ratio 1 : 2.

Example – 4

Find the area of the parallelogram formed by the lines

\[\begin{align}&{a_1}x + {b_1}y + {c_1} = 0 \; ; \; {a_1}x + {b_1}y + {d_1} = 0  \\ & {a_2}x + {b_2}y + {c_2} = 0 \; ; \; {a_2}x + {b_2}y + {d_2} = 0  \\ \end{align} \]

Find also the condition for this parallelogram to be a rhombus.

Solution:

We first consider a little geometry for this parallelogram. Let the parallelogram have sides a and b and let the perpendicular distances between its opposite sides be p1 and p2:

Then, the area A of the parallelogram is

\[\begin{align}  A = ab\sin \alpha  &= a{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;(\because \,\,\,{p_1} = b\sin \alpha ) \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{{p_2}}}{{\sin \alpha }} \cdot {p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because \,\,\,{p_2} = a\sin \alpha )\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{{p_1}{p_2}}}{{\sin \alpha }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)  \\ \end{align} \]

Thus, the area of the parallelogram can be expressed using the perpendicular distances between its opposite sides rather than using the length of the sides. This is good for us since we already know how to evaluate the perpendicular distance between two parallel lines.

\[ \Rightarrow \quad {p_1} = \frac{{\left| {{c_1} - {d_1}} \right|}}{{\sqrt {a_1^2 + b_1^2} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

\[ \Rightarrow \quad {p_2} = \frac{{\left| {{c_2} - {d_2}} \right|}}{{\sqrt {a_2^2 + b_2^2} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\]

Also,

\[\begin{align}\tan \alpha &= \left| {\frac{{\left( { - \frac{{{a_1}}}{{{b_1}}}} \right) - \left( { - \frac{{{a_2}}}{{{b_2}}}} \right)}}{{1 + \frac{{{a_1}{a_2}}}{{{b_1}{b_2}}}}}} \right|  \\  \,\,\,\,\,\,\,\,\,\,\,\,\, &= \left| {\frac{{{a_1}{b_2} - {a_2}{b_1}}}{{{a_1}{a_2} + {b_1}{b_2}}}} \right|  \\ \end{align} \]

so that

\[\sin \alpha  = \frac{{\left| {{a_1}{b_2} - {a_2}{b_1}} \right|}}{{\sqrt {(a_1^2 + b_1^2)(a_2^2 + b_2^2)} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 4 \right)\]

Using (2), (3) and (4) in (1), we have

\[A = \left| {\frac{{({c_1} - {d_1})({c_2} - {d_2})}}{{({a_1}{b_2} - {a_2}{b_1})}}} \right|\]

Now, the parallelogram is a rhombus if its diagonals are perpendicular, which also means that the distances between its opposite sides are equal, i.e.

\[\begin{align} &{p_1} = {p_2}  \\   \Rightarrow  \qquad&\frac{{\left| {{c_1} - {d_1}} \right|}}{{\sqrt {a_1^2 + b_1^2} }} = \frac{{\left| {{c_2} - {d_2}} \right|}}{{\sqrt {a_2^2 + b_2^2} }}  \\ \end{align}\]

Download practice questions along with solutions for FREE:
Straight Lines
grade 11 | Answers Set 2
Straight Lines
grade 11 | Questions Set 1
Straight Lines
grade 11 | Answers Set 1
Straight Lines
grade 11 | Questions Set 2
Download practice questions along with solutions for FREE:
Straight Lines
grade 11 | Answers Set 2
Straight Lines
grade 11 | Questions Set 1
Straight Lines
grade 11 | Answers Set 1
Straight Lines
grade 11 | Questions Set 2
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