Question

a
The point on the line 3x + 4y = 5 which is equidistant from (1, 2) and (3, 4) is (alfa, beta) then alfa/ beta
IS​

Answer 1

Step-by-step explanation:

Let's take a general point on the line 3x+4y−5=0, which is (x,45−3x)

According to given condition, this point is equidistant from points (1,2) and (3,4)

Using distance formula, by equalizing their distances we get,

⇒(2−45−3x)2+(1−x)2=(4−45−3x)2+(3−x)2

⇒(2)2+(45−3x)2−(5−3x)+12+x2−2x=(4)2+(45−3x)2−2(5−3x)+32+x2−6x

⇒x=15

⇒ Hecne y=45−3(15)=−10

So the point is (15,−10)

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Answer 2

ANSWER

ANSWERLet's take a general point on the line 3x+4y−5=0, which is (x,45−3x)

ANSWERLet's take a general point on the line 3x+4y−5=0, which is (x,45−3x)According to given condition, this point is equidistant from points (1,2) and (3,4)

ANSWERLet's take a general point on the line 3x+4y−5=0, which is (x,45−3x)According to given condition, this point is equidistant from points (1,2) and (3,4)Using distance formula, by equalizing their distances we get,

ANSWERLet's take a general point on the line 3x+4y−5=0, which is (x,45−3x)According to given condition, this point is equidistant from points (1,2) and (3,4)Using distance formula, by equalizing their distances we get,⇒(2−45−3x)2+(1−x)2=(4−45−3x)2+(3−x)2

⇒(2)2+(45−3x)2−(5−3x)+12+x2−2x=(4)2+(45−3x)2−2(5−3x)+32+x2−6x

⇒(2)2+(45−3x)2−(5−3x)+12+x2−2x=(4)2+(45−3x)2−2(5−3x)+32+x2−6x⇒x=15

⇒(2)2+(45−3x)2−(5−3x)+12+x2−2x=(4)2+(45−3x)2−2(5−3x)+32+x2−6x⇒x=15⇒ Hence y=45−3(15)=−10

e y=45−3(15)=−10So the point is (15,−10)