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दर्शाइए कि रेखाओ `y =m _(1 ) x +c _(1 ),y =m _(2 ) x +c _(2 )` और `x =0 ` से बने त्रिभुज का क्षेत्रफल `((c_(1)-c_(2))^(2)) /(2|m_(1)0m_(2)|)`है

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दी गयी रेखाएँ है
`y=m_(1)x+c_(1 )`
`y=m_(2)x+c_(2)`
तथा `x=0`
समीकरण (i ) तथा (ii ) को हल करने पर ,
`x=0`तथा `y=c_(1)`
अर्थात बिंदु ` P (0 , c _(1 ))` प्राप्त होता है ।
समीकरण (ii ) तथा (iii ) को हल करने पर प्राप्त बिंदु `Q (0 ,c _(2 ))` है ।
अब : समीकरण (i ) तथा (ii ) को हल करने पर ,
`x=(c_(2)-c_(1))/(m_(1)-m_(2))`तथा `y=(m_(1)c_(2)-m_(2)c_(1))/(m_(1)-m_(2))`
अर्थात तीसरा बिंदु `R ((C_(2)-c_(1))/(m_(1(-m_(2)),(m_(1)C_(2)-m_(2)c_(1))/(m_(1)-m_(2)))` है ।
इन तीनो बिन्दुओ से प्राप्त त्रिभुज का क्षेत्रफल
`=(1)/(2)|0.((m_(1)C_(1)-m_(2)c_(1))/(m_(1)-m_(2))-c_(2))+(c_(2)-c_(1))/(m_(1)-m_(2))(c_(2)-c_(1))+0.(C_(1)-(m_(1C_(2)-m_(2)c_(1))/(m_(1)-m_(2)))|`

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