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Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4units.

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As an experienced tutor registered on UrbanPro, I'd be delighted to help you with this math problem. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services. To find the points on the x-axis that are a distance of 4 units from the line equation (x/3)+(y/4)=1(x/3)+(y/4)=1,...
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As an experienced tutor registered on UrbanPro, I'd be delighted to help you with this math problem. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services. To find the points on the x-axis that are a distance of 4 units from the line equation (x/3)+(y/4)=1(x/3)+(y/4)=1, we can use the distance formula between a point (x,y)(x,y) and a line. The distance between a point (x0,y0)(x0,y0) and a line Ax+By+C=0Ax+By+C=0 is given by: Distance=∣Ax0+By0+C∣A2+B2Distance=A2+B2 ∣Ax0+By0+C∣ In our case, the line equation is (x/3)+(y/4)=1(x/3)+(y/4)=1, which can be rewritten in the standard form as 4x+3y−12=04x+3y−12=0. So, A=4A=4, B=3B=3, and C=−12C=−12. Now, we're looking for points on the x-axis, so their y-coordinate is 0. Let's denote the x-coordinate of such a point as x0x0. Plugging these values into the distance formula: ∣4x0+3(0)−12∣42+32=442+32 ∣4x0+3(0)−12∣=4 ∣4x0−12∣5=45∣4x0−12∣=4 Now, we can solve for x0x0: ∣4x0−12∣=20∣4x0−12∣=20 4x0−12=±204x0−12=±20 Solving for x0x0: When 4x0−12=204x0−12=20: 4x0=324x0=32 x0=8x0=8 When 4x0−12=−204x0−12=−20: 4x0=−84x0=−8 x0=−2x0=−2 So, the points on the x-axis that are 4 units away from the given line are x=8x=8 and x=−2x=−2. read less
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